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GRE Subject Test: Chemistry : Analyzing Solids

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Example Question #51 : Analytical Chemistry

Calculate the molar solubility of $Co(OH)_{2}$ with $Ksp = 1.3\times 10^{-15}$ if enough NaOH is added to raise the pH of the solution to pH 12.

Possible Answers:

$1.3\times10^{-13}\textup{ M}$

$0.5421\textup{ M}$

$4.11\textup{ M}$

$2.3\times10^{-3}\textup{ M}$

$3.1\times10^{-3}\textup{ M}$

Correct answer:

$1.3\times10^{-13}\textup{ M}$

Explanation:

The equation for the dissolution of $Co(OH)_{2}$ in water is:

$Co(OH)_{2(s)}\rightleftharpoons Co^{2+}(aq)+2OH^{-}(aq)$

The $K_{sp}$ for the above equation is:

$Ksp=[Co^{2+}][OH^{-}]^{2}=1.3\times10^{-15}$

Due to the solubility of $Co(OH)_{2}$ in water, the concentration of $Co^{2+}$ can be set to:

$[Co^{2+}]=X$

Therefore the concentration of $OH^{-}$ should be double that of $Co^{2+}$ :

$[OH^{-}]=2X$

The solution was adjusted to pH 12 using NaOH , therefore:

pH +pOH=14

pOH=14-12=2

$[OH^{-}]=10^{-pOH}=10^{-2}=1.0\times10^{-2}$

Using an ICE table to process the data:

Screen shot 2015 11 13 at 4.39.13 pm

Plugging these variables into the $K_{sp}$ equation gives:

$Ksp=[X]\times [0.01+2X^{2}]=1.3\times10^{-15}$

$2X^{2}<< 0.01$

Therefore, the $K_{sp}$ equation can be approximated to:

$Ksp=[X]\times [0.01]=1.3\times10^{-15}$

$[X]=\frac{1.3\times10^{-15}}{[0.01]}=1.3\times10^{-13}M$

Therefore, the solubility of $Co(OH)_{2}$ in water is $1.3\times10^{-13}M$ .

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Example Question #51 : Analytical Chemistry

$Pb(OH)_{2}$ has a $K_{sp}$ equal to $1.2\times10^{-15}$ . What would be the numerical expression used to determine the molar solubility (S) of $Pb(OH)_{2}$ ?

Possible Answers:

$\sqrt[2]{1.2\times10^{-15}}$

$\sqrt[3]{6.0\times10^{-16}}$

$\sqrt[3]{4.0\times10^{-16}}$

$\sqrt[2]{4.0\times10^{-16}}$

Correct answer:

$\sqrt[3]{6.0\times10^{-16}}$

Explanation:

The equation for the dissolution of $Pb(OH)_{2}$ in water is below:

$Pb(OH)_2_{(s)}\rightleftharpoons\ Al^{2+}_{(aq)}\ +\ 2OH^{-}_{(aq)}$

The Ksp expression for the above chemical equation is:

$K_{sp}=S\times\2S^{2}=2S^{3}$

Rearranging this equation to solve for solubility (S) gives:

$\frac{K_{sp}}{2}=S^{3}$

$\sqrt[3]{\frac{K_{sp}}{2}}=S$

Plugging the value for Ksp into the equation gives:

$S=\sqrt[3]{\frac{1.2\times10^{-15}}{2}}=\sqrt[3]{6.0\times10^{-16}}$

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Example Question #51 : Analytical Chemistry

Calculate the molar solubility of $Mn(OH)_{2}$ with a $K_{sp} = 1.6\times 10^{-13}$ if enough NaOH is added to raise the pH of the solution to pH 10.

Possible Answers:

$7.9\times10^{-3}\textup{ M}$

$0.1992\textup{ M}$

$1.2\times10^{-7}\textup{ M}$

$3.2\textup{ M}$

$1.6\times10^{-9}\textup{ M}$

Correct answer:

$1.6\times10^{-9}\textup{ M}$

Explanation:

The equation for the dissolution of $Mn(OH)_{2}$ in water is below: $Mn(OH)_{2(s)}\rightleftharpoons Mn^{2+}(aq)+2OH^{-}(aq)$

The $K_{sp}$ for the above equation is:

$Ksp=[Mn^{2+}][OH^{-}]^{2}=1.3\times10^{-15}$

Due to the solubility of $Mn(OH)_{2}$ in water, the concentration of $Mn^{2+}$ can be set to:

$[Mn^{2+}]=X$

Therefore the concentration of $OH^{-}$ should be double that of $Mn^{2+}$ :

$[OH^{-}]=2X$

The solution was adjusted to pH 10 using NaOH , therefore:

pH +pOH=14

pOH=14-10=4

$[OH^{-}]=10^{-pOH}=10^{-4}=1.0\times10^{-4}$

Using an ICE table to process the data:

Screen shot 2015 11 13 at 4.35.54 pm

Plugging into the $K_{sp}$ equation gives:

$Ksp=[X]\times [0.0001+2X^{2}]=1.6\times10^{-13}$

$2X^{2}<< 0.0001$

Therefore, the $K_{sp}$ equation can be approximated to:

$Ksp=[X]\times [0.0001]=1.6\times10^{-13}$

$[X]=\frac{1.6\times10^{-13}}{[0.0001]}=1.6\times10^{-9}M$

Therefore, the solubility of $Mn(OH)_{2}$ in water is $1.6\times10^{-9}M$ .

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Example Question #52 : Analytical Chemistry

Calculate the molar solubility of PbF_2 with $K_{sp} = 3.6\times 10^{-8}$ in a solution containing $0.01\textup{ M}$ NaF .

Possible Answers:

$2.1\times10^{-3}\textup{ M}$

$5.8\times10^{-4}\textup{ M}$

$1.0\times10^{-2}\textup{ M}$

$3.6\times10^{-6}\textup{ M}$

$5.1\times10^{-8}\textup{ M}$

Correct answer:

$3.6\times10^{-6}\textup{ M}$

Explanation:

The equation for the dissolution of PbF_2 in water is below: $PbF_{2(s)}\rightleftharpoons Pb^{2+}(aq)+2F^{-}(aq)$

The $K_{sp}$ for the above equation is:

$Ksp=[Pb^{2+}][F^{-}]^{2}=3.6\times10^{-8}$

Due to the solubility of PbF_2 in water, the concentration of $Pb^{2+}$ can be set to be:

$[Pb^{2+}]=X$

Therefore F^- concentration will be double:

$F^{-}=2X$

Using an ICE table to process the data:

Screen shot 2015 11 09 at 9.29.39 pm

Plugging these variable into the $K_{sp}$ equation gives:

$Ksp=[X]\times [0.010+2X^{2}]=3.6\times10^{-8}$

$2X^{2}<< 0.010$

Therefore, the $K_{sp}$ equation can be approximated to:

$Ksp=[X]\times [0.010]=3.6\times10^{-8}$

$[X]=\frac{3.6\times10^{-8}}{[0.010]}=3.6\times10^{-6}M$

Therefore, the solubility of PbF_2 in water is $3.6\times10^{-6}M$ .

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Example Question #51 : Analytical Chemistry

Calculate the molar solubility of PbF_2 with $K_{sp} = 3.6\times 10^{-8}$ in a solution containing $0.10\textup{ M}$ $Pb(NO_{3})_{2}$ .

Possible Answers:

$3.1\times10^{-7}\textup{ M}$

$2.0\times10^{-3}\textup{ M}$

$1.2\times10^{-2}\textup{ M}$

$4.2\times10^{-4}\textup{ M}$

$0.111\textup{ M}$

Correct answer:

$4.2\times10^{-4}\textup{ M}$

Explanation:

The equation for the dissolution of PbF_2 in water is below: $PbF_{2(s)}\rightleftharpoons Pb^{2+}(aq)+2F^{-}(aq)$

The $K_{sp}$ for the above equation is:

$Ksp=[Pb^{2+}][F^{-}]^{2}=3.6\times10^{-8}$

Due to the solubility of PbF_2 in water, taking into account the concentration of $Pb^{2+}$ from PbF_2 and $Pb(NO_{3})_{2}$ can be set to:

$[Pb^{2+}]=X+0.10$

The F^- concentration will be double the concentration of the $Pb^{2+}$ coming from PbF_2 :

$F^{-}=2X$

Using an ICE table to process the data:

Screen shot 2015 11 09 at 10.02.00 pm

Plugging these variables into the $K_{sp}$ equation gives:

$Ksp= [0.10+X]\times [2X^{2}]=3.6\times10^{-8}$

X<< 0.10

Therefore, the $K_{sp}$ equation can be approximated to:

$Ksp=[0.10]\times [2X]^{2}=3.6\times10^{-8}$

$[2X]^{2}=\frac{3.6\times10^{-8}}{[0.10]}$

$[X]^{2}=\frac{3.6\times10^{-8}}{[0.10]\times 2}$

$[X]=\sqrt{\frac{3.6\times10^{-8}}{[0.10]\times 2}}=4.2\times10^{-4}\textup{ M}$

Therefore, the solubility of PbF_2 in water is $4.2\times10^{-4}\textup{ M}$ .

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Example Question #53 : Analytical Chemistry

$Al(OH)_{3}$ has a $K_{sp}$ equal to $1.3\times10^{-33}$ . What would be the numerical expression used to determine the molar solubility (S) of $Al(OH)_{3}$ ?

Possible Answers:

$\sqrt[4]{6.5\times10^{-34}}$

$\sqrt[4]{4.3\times10^{-34}}$

$\sqrt[3]{4.3\times10^{-33}}$

$\sqrt[3]{2.2\times10^{-33}}$

Correct answer:

$\sqrt[4]{4.3\times10^{-34}}$

Explanation:

The equation for the dissolution of $Al(OH)_{3}$ in water is below:

$Al(OH)_3_{(s)}\rightleftharpoons\ Al^{3+}_{(aq)}\ +\ 3OH^{-}_{(aq)}$

The Ksp expression for the above chemical equation is:

$K_{sp}=S\times\3S^{3}=3S^{4}$

Rearranging this equation to solve for solubility (S) gives:

$\frac{K_{sp}}{3}=S^{4}$

$\sqrt[4]{\frac{K_{sp}}{3}}=S$

Plugging the value for Ksp into the equation gives:

$S=\sqrt[4]{\frac{1.3\times10^{-33}}{3}}=\sqrt[4]{4.3\times10^{-34}}$

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Example Question #51 : Analytical Chemistry

For the titration of $0.001 M\ Ag^{2+}$ with 0.150 L solution of $Cl^-_{(aq)}$ , the end point occurs upon addition of 0.200 L of the $Ag^{2+}$ solution. Determine the initial concentration of chloride that was present in the original solution.

Possible Answers:

$0.0026\ M$

$1.321\ M$

$0.063\ M$

$0.0013\ M$

Correct answer:

$0.0013\ M$

Explanation:

The equivalence point is when enough titrant has been added so that the number of moles of titrant equals the number of moles of analyte. We must first determine the number of moles of silver ions at the equivalence point.

At the equivalence point the following occurs in solution:

$moles\ of\ Ag^{2+}=moles\ of\ Cl^{-}$

The number of moles of silver ions at equivalence point is calculated as follows:

$0.200\ L\times \frac{0.001\ moles}{L}=2.0\times 10^{-4}\ moles\ Ag^{2+}$

$Molarity=\frac{moles\ of\ solute}{Liters\ of\ solution}$

$Liters\ of\ solution= Original\ volume\ of\ Cl^{-}\ solution$

Plugging these values into the equation gives:

$\frac{2\times10^{-4}\ moles}{0.150\ L}=0.0013\ M\ Cl^{-}$

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Example Question #13 : Solubility And Solution Equilibrium

How many moles of fluoride ions are present in of a completely saturated solution of lead (II) fluoride?

$PbF_2(s)\rightarrow Pb^{2+}(aq)+2F^-(aq)$

$K_{sp}=3.7*10^{-8}$

Possible Answers:

$2.1*10^{-3}mol$

$8.4*10^{-3}mol$

$4.2*10^{-3}mol$

$1.5*10^{-7}mol$

$7.4*10^{-8}mol$

Correct answer:

$4.2*10^{-3}mol$

Explanation:

Recall that the solubility product constant is given by the equation below, for a reaction in the following format.

$aA\rightarrow cC+dD$

$K_{sp}=[C]^c[D]^d$

Using our balanced reaction, we can find the solubility product equation for the dissociation of lead (II) fluoride.

$PbF_2(s)\rightarrow Pb^{2+}(aq)+2F^-(aq)$

$K_{sp}=[Pb^{2+}][F^-]^2$

For each molecule of lead (II) fluoride that dissolves, it produces one lead ion and two fluoride ions. We can conclude that the concentration of fluoride ions in solution will be twice the concentration of lead ions.

$[Pb^{2+}] = x$

[F^-]=2x

Use these variables in the solubility product equation, along with the given value from the question.

$3.7*10^{-8}=(x) (2x)^2$

Now we can solve for the value of $\small x$ .

$3.7*10^{-8}=4x^3$

$x = \sqrt[3]{\frac{3.7*10^{-8}}{4}}=2.1*10^{-3}$

Remember that this value is equal to the concentration of lead ions in solution and half the concentration of fluoride ions in solution.

$[Pb^{2+}]=x=2.1*10^{-3}M$

$[F^-]=2x=2(2.1*10^{-3}M)=4.2*10^{-3}M$

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Example Question #52 : Analytical Chemistry

Calculate the molar solubility of $Co(OH)_{2}$ which has a $Ksp = 1.3\times 10^{-15}$ .

Possible Answers:

$6.36\times10^{-6}\textup{ M}$

$6.88\times10^{-6}M$

$2.09\times10^{-3}\textup{ M}$

$8.66\times10^{-6}\textup{ M}$

$3.6\times10^{-6}\textup{ M}$

Correct answer:

$6.88\times10^{-6}M$

Explanation:

The equation for the dissolution of $Co(OH)_{2}$ in water is:

$Co(OH)_{2(s)}\rightleftharpoons Co^{2+}(aq)+2OH^{-}(aq)$

The $K_{sp}$ for the above equation is:

$Ksp=[Co^{2+}][OH^{-}]^{2}=1.3\times10^{-15}$

Due to the solubility of $Co(OH)_{2}$ in water, the concentration of $Co^{2+}$ can be set to:

$[Co^{2+}]=x$

Therefore the concentration of $OH^{-}$ should be double that of $Co^{2+}$ :

$[OH^{-}]=2x$

Plugging these variables into the $K_{sp}$ equation gives:

$Ksp=x\cdot (2x)^{2}$

Ksp=4x3

Now, we can solve for the x-variable by setting this value equal to the given $K_{sp}$ .

$4^{3}=1.3\times10^{-15}$

$\frac{1.3\times 10^{-15}}{4}=x^{3}$

$x=\sqrt[3]{\frac{1.3\times 10^{-15}}{4}}=6.87534\times10^{-6}$

Therefore, the solubility of $Co(OH)_{2}$ in water is $6.88\times10^{-6}M$ .

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Example Question #1 : Mass Spectroscopy

Determining the molecular ion peak (parent peak) in mass spectroscopy allows you to determine what characteristic of a mystery molecule?

Possible Answers:

Molecular charge

Nuclear charge

Functional groups

Molecular weight

Correct answer:

Molecular weight

Explanation:

The molecular ion peak is determined using mass spectrometry. The parent peak is formed when a mystery molecule does not fragment, and simply loses an electron. This means that the mass to charge ratio of this peak will allow us to determine the molecular weight of the compound.

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GRE Subject Test: Chemistry : Analyzing Solids

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #51 : Analytical Chemistry

Example Question #51 : Analytical Chemistry

Example Question #51 : Analytical Chemistry

Example Question #52 : Analytical Chemistry

Example Question #51 : Analytical Chemistry

Example Question #53 : Analytical Chemistry

Example Question #51 : Analytical Chemistry

Example Question #13 : Solubility And Solution Equilibrium

Example Question #52 : Analytical Chemistry

Example Question #1 : Mass Spectroscopy

All GRE Subject Test: Chemistry Resources

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